The tires of a 1500 kg car are 0.600 m in diameter and the coefficients of friction with the road surface are Âµs = 0.800 and Âµk = 0.600. Suppose the car has a disk brake system. Each wheel is slowed by the frictional force between a single brake pad and the disk-shaped rotor. On this particular car, the brake pad comes into contact with the rotor at an average distance of 18.5 cm from the axis. The coefficients of friction between the brake pad and the disk are µs = 0.588 and µk = 0.490. Calculate the normal force that must be applied to the rotor such that the car slows as quickly as possible.
Torque = Fd
Friction = n* coefficiant
The Attempt at a Solution
I'm not very sure which values to use. All I did was find the torque required and find the amount of force to apply
0.8 * 1500/4 * 9.8 = 0.185 * F * 0.588
8.11, 1.35 and 7.30 kN are some of the answers that I submitted and are all wrong.